Question

Решите неопределенный интеграл

Answer 1

$\displaystyle \int\limits {\dfrac{\ln \text{tg} \, x}{\sin x \cos x} } \, dx = \begin{vmatrix}\ln \text{tg} \, x = t\\ \dfrac{dx}{\sin x \cos x} = dt \\ \end{vmatrix} = \int\limits {t} \, dt = \dfrac{t^{2}}{2} + C= \dfrac{\ln^{2}\text{tg} \, x}{2} +C$

Как нашли $\dfrac{dx}{\sin x \cos x}:$

$\dfrac{d(\ln \text{tg} \, x)}{dx} = (\ln \text{tg} \, x)' = \dfrac{1}{\text{tg} \, x} \cdot (\text{tg} \, x)' = \dfrac{1}{\dfrac{\sin x}{\cos x} } \cdot \dfrac{1}{\cos^{2}x} = \dfrac{1}{\sin x \cos x}$

Ответ: $\dfrac{\ln^{2}\text{tg} \, x}{2} +C$